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Question Number 75248 by cesar.marval.larez@gmail.com last updated on 08/Dec/19 Commented by mathmax by abdo last updated on 14/Dec/19 $$\frac{{d}}{{dx}}\left(\frac{{e}^{{x}} }{\mathrm{1}+{x}}\right)\:=\frac{{e}^{{x}} \left(\mathrm{1}+{x}\right)−{e}^{{x}} ×\mathrm{1}}{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\:=\frac{\mathrm{xe}^{\mathrm{x}} }{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}} }\:\Rightarrow…
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